Let's go through the problems one by one from the file you uploaded. I'll start with the second one, where you are asked to verify Rolle's theorem for $f(x) = x(x + 3)e^{-x/2}$ on the interval $[-3, 0]$.

Rolle's Theorem:

Rolle's Theorem states that if a function $f(x)$ is:

1. Continuous on a closed interval $[a, b]$,
2. Differentiable on the open interval $(a, b)$,
3. $f(a) = f(b)$,

then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

Step 1: Check continuity and differentiability

The function $f(x) = x(x + 3)e^{-x/2}$ is a product of polynomials and an exponential function, both of which are continuous and differentiable on any interval. Thus, $f(x)$ is continuous and differentiable on $[-3, 0]$.

Step 2: Check the values of $f(-3)$ and $f(0)$

• $f(-3) = (-3)(-3 + 3)e^{3/2} = 0$,
• $f(0) = (0)(0 + 3)e^{0} = 0$.

Since $f(-3) = f(0) = 0$, the condition $f(a) = f(b)$ is satisfied.

Step 3: Find $f'(x)$

We differentiate $f(x) = x(x + 3)e^{-x/2}$ using the product and chain rule:

$f'(x) = \frac{d}{dx} \left( x(x + 3) e^{-x/2} \right)$ Let $g(x) = x(x + 3)$ and $h(x) = e^{-x/2}$.

By the product rule: $f'(x) = g'(x) h(x) + g(x) h'(x)$

First, find the derivatives of $g(x)$ and $h(x)$: $g'(x) = (2x + 3), \quad h'(x) = \frac{d}{dx} \left( e^{-x/2} \right) = -\frac{1}{2} e^{-x/2}$

Now substitute these into the product rule: $f'(x) = (2x + 3)e^{-x/2} + x(x + 3)\left( -\frac{1}{2}e^{-x/2} \right)$

Simplify: $f'(x) = e^{-x/2} \left( (2x + 3) - \frac{1}{2}x(x + 3) \right)$ $f'(x) = e^{-x/2} \left( (2x + 3) - \frac{1}{2}(x^2 + 3x) \right)$ $f'(x) = e^{-x/2} \left( 2x + 3 - \frac{1}{2}x^2 - \frac{3}{2}x \right)$ $f'(x) = e^{-x/2} \left( -\frac{1}{2}x^2 + \frac{1}{2}x + 3 \right)$

Step 4: Solve $f'(x) = 0$

Set $f'(x) = 0$: $-\frac{1}{2}x^2 + \frac{1}{2}x + 3 = 0$

Multiply the equation by $-2$ to simplify: $x^2 - x - 6 = 0$

Factor the quadratic: $(x - 3)(x + 2) = 0$

Thus, $x = 3$ or $x = -2$.

Step 5: Check if $x = -2$ is in the interval $(-3, 0)$

Since $x = -2$ lies in the interval $(-3, 0)$, there is a value $c = -2$ such that $f'(c) = 0$. Therefore, Rolle's theorem is verified for $f(x)$ on $[-3, 0]$.

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Would you like further details on any of the steps?

Here are five follow-up questions you may find helpful:

1. How can Rolle's theorem be applied to other types of functions?
2. What is the geometric interpretation of Rolle's theorem?
3. How does Rolle's theorem relate to the Mean Value Theorem?
4. Can you provide an example where Rolle's theorem does not apply?
5. How would the verification change if the interval were different?

Tip: When verifying Rolle's or Mean Value Theorem, always check continuity, differentiability, and equal boundary values before finding critical points.